208 IX ANALYTIC FUNCTIONS containing [0, 1]. Let A be the closed subset of the interval [0, 1] consisting of the t such that h(s) = 0 for 0 ^ s < t; by assumption there is an open neighborhood of 0 in [0, 1] which is contained in A, hence the l.u.b. p of A is certainly >0; we will prove that p = 1, which will establish (9.4.1). Note first that h(f) = 0 for 0 < t < p, hence by continuity h(p) = 0; as h is analytic at the point p, there is a power series in t - p, which converges for \t - p\ < a, with a > 0 and whose sum is equal to h(t) for \t — p\